|
In more detail: If i have two soda cans, both are cooled to exactly 4 degrees celsius, And i put one in a 25 degrees room, and the other next to an AC vent set to 16 degrees. After three minutes, which one should be colder than the other and why? Edit: To clarify - if I have a cold soda can, should I place it near the AC vent or not (if I like my drink cold)? Which location will cause faster heating? |
|||||||||||
|
|
At the level you are asking the question it should be obvious that the closer to the cold output of the AC the can is sitting the cooler it will remain, since next to the vent the temperature will be the coldest and not yet mixed with the air in the room, a "refrigerator" set up. The soda can when coming out of the refrigerator will be 4C but the air coming right out of the AC is much warmer, 16C, the close can will heat up to that in equilibrium. If the other can is at 25C it will reach an equilibrium at that temperature and will be warmer.The argument still holds for 3 minutes, the one at 25C will be incrementally warmer than the one at 16C. If you are interested in the physics basis of this and real numbers, here is a chapter on heat transfer.: conduction, convection and radiation. Boundary conditions are necessary to get real numbers from differential equations. |
|||||||
|
|
The answer for the question as posed is easy: It could be either This applies for the initial rate of heating. Of course, over time, the story is quite different. The temperature of the one in front of the AC could initially become higher than the one in the static room air. However, given sufficient time, of course the can in front of the AC will be the cooler can and will remain that way. Simple Newton's law of cooling: $$\frac{dT_{can}}{dt} = C_{air} \left( T_{air} - T_{can} \right)$$ $$\frac{dT_{can}}{dt} = C_{AC} \left( T_{AC} - T_{can} \right)$$ The solution to both of these is simple. Written for both cases, they are: $$T_{can}(t) = T_{air} - ( T_{air}-T_{can}(0)) exp(-C_{air} t)$$ $$T_{can}(t) = T_{AC} - ( T_{AC}-T_{can}(0)) exp(-C_{AC} t)$$ I can not make any statements about the exact values, and I don't think that's likely to be valuable for this exercise, but allow me to make simple relative statements. $$ T_{air} > T_{AC}$$ $$C_{air} < C_{AC} $$ It's taken as a given that the initial can temperature is the same for both cases. Should the above inequalities have the same direction, then the problem would have an absolute answer. One can would be hotter at all times, $t>0$. In the problem presented, the signs are different leading to two possibilities, which is that the AC case initially leads to a higher temperature, or that the AC case is always of lower temperature. This can be formalized by the following inequality. $$C_{AC} \left( T_{AC} - T_{can}(0) \right) \stackrel{?}{\le} C_{air} \left( T_{air} - T_{can}(0) \right)$$ If the above inequality is true, then the can in front of the AC is always a lower temperature. If it is false, then the can in front of the AC initially becomes hotter, then the other can becomes hotter. |
|||
|
Tricky. The 25C room is a static environment, whereas in the 16C AC-blowing room, the can environment is forced to be at 16C. After 3 minutes, which is a short amount of time, I think we need to take into account many factors, like - among others - if there is moisture in the air. The can that is blown on by the AC will have more thermodynamics of condensing-evaporating humidity than the one that is not blown on. On the other hand the can that is not blown on, will cool the surroundings and depending on how it is placed in the room (suspended? on a surface? which material is the surface made of?) will cool the air around it, dampening temperature gradient between the liquid at 4C and the bulk of the room at 25C. But that cold air will have a different density, so convection comes into play, etc, etc. Please simplify your problem or make an experiment. |
|||
|
|
|
The air around the soda can effects how fast it transfers heat. If the temperature difference is low it will change gradually because the actual energy it takes to transfer energy is constant. waters specific heat is 4181.3 J/(kg·K), water has the second highest specific heat capacity. This means it takes a certain amount of energy to transfer. The lower the temperature the slower the energy transfered is. The can in the cold air should remain colder from the exprience from labs i've done with it in the past. |
|||||||||||
|
